Chemistry question-electrochemistry

Question

Cadmium sulfide(CdS) is used in some semiconductor applications. Calculate the value of the solubility product constant(Ksp) for CdS given the following standard reduction potentials:

CdS 2e- ---> Cd S^2- E= -1.21 V

Cd^2 2e- -->Cd E=-0.402 V

Answer 1

if:

Cd+2   &    2e- -->   Cd .................E= - 0.402 V

then:

Cd -->   Cd+2   &    2e- .................E= + 0.402 V

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combining

Cd -->   Cd+2   &    2e- .................E= + 0.402 V

CdS  &   2e- ---> Cd   &    S^2-    ...........E= -1.21 V

gives:

CdS --> Cd+2  &  S-2  ....Eo = - 0.808 volts

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log K = [(Eo)  (n)] / 0.0592

log K = [(-0.808)  (2)] / 0.0592

log K =-- 27 297

your answer is: K = 5.0e-28